Question

A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water. What is his speed just before striking the water

Answer 1

Answer:

14.05m/s

Explanation:

We are given that

Horizontal speed of diver=$v_x=1.2m/s$

Distance, y=-10 m

It is taken as negative because the diver goes downward.

There is no air resistance therefore, there is  acceleration due to gravity .

We know that

$g=-9.8 m/s^2$

Initial vertical velocity, $u_y=0$

$v^2-u^2=2as$

Using the formula

$v^2_y-0=2\times (-9.8)(-10)$

$v_y=\sqrt{2(-9.8)(-10)}$

$v_y=14 m/s$

$v=\sqrt{v^2_x+v^2_y}$

Using the formula

$v=\sqrt{(1.2)^2+(14)^2}=14.05m/s$

Hence, the speed of diver before just striking the water=14.05m/s